$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
L &= 1.00 text{ m} sqrt{1 – dfrac{(0.999 c)^2}{c^2}} \
L &= 0.0447 m
end{align*}
$$

The length of the meterstick contracts as speed increases, if the speed is 0.999 c, the length of the meterstick is $boxed{text{0.0447 m}}$

$L = 2.00$ m

$v = 0.995 c$

$textbf{Find:}$ Length of the wood at rest, $L_o$

Using the length contraction formula, we can calculate $L_o$ as follows:

$$
begin{align*}
L_o &= dfrac{L}{sqrt{1 – dfrac{v^2}{c^2}}} \
L_o &= dfrac{2.00 text{ m}}{sqrt{1 – dfrac{(0.995c)^2}{c^2}}} \
L_o &= boxed{20 text{ m}}
end{align*}
$$

The length contraction formula is given by

$$
begin{align}
L=L_0 sqrt{1-frac{v^2}{c^2}}
end{align}
$$

where $L$ is the length of the (one dimensional) object in a lab frame of reference, $L_0$ is the length of the object in a frame of reference in which the object is at rest, $v$ is the speed of the moving frame of reference and $c$ is the speed of light in vacuum.

The proper length of the meterstick $L_0$ is

$$
begin{align*}
L_0=1 text{m}
end{align*}
$$

The length of the meterstick in the laboratory frame of reference $L$ is

$$
begin{align*}
L=0.5 text{m}
end{align*}
$$

By using Eq. (1) we can express the speed of the meterstick $v$ as

$$
begin{align*}
frac{v^2}{c^2}&=1-frac{L^2}{L_0^2}\
&=1-frac{(0.5 text{m})^2}{(1 text{m})^2}\
&=0.75
end{align*}
$$

Thus, the speed of the meterstick $v$ is

$$
begin{align*}
boxed{v}&=sqrt{0.75}c\
&=boxed{0.86c}
end{align*}
$$